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Popular Trigonometria >

cot((3pi+x)/2)<= 1

Solução

cot (\frac{3 π + x}{2}) \leq 1

Solução

- \frac{5 π}{2} + 2 πn \leq x < - π + 2 πn

Notação de intervalo

[- \frac{5 π}{2} + 2 πn, - π + 2 πn)

Decimal

- 7.85398 \dots + 2 πn \leq x < - 3.14159 \dots + 2 πn

Passos da solução

cot (\frac{3 π + x}{2}) \leq 1

cot (x) \leq a

então

arccot (a) + πn \leq x < π + πn

arccot (1) + πn \leq \frac{3 π + x}{2} < π + πn

a \leq u < b

então

a \leq u and u < b

arccot (1) + πn \leq \frac{3 π + x}{2} and \frac{3 π + x}{2} < π + πn

arccot (1) + πn \leq \frac{3 π + x}{2} : x \geq - \frac{5 π}{2} + 2 πn

arccot (1) + πn \leq \frac{3 π + x}{2}

Trocar lados

\frac{3 π + x}{2} \geq arccot (1) + πn

Simplificar

arccot (1) + πn : \frac{π}{4} + πn

arccot (1) + πn

Utilizar a seguinte identidade trivial:

arccot (1) = \frac{π}{4}

x - 3 - 1 - \frac{3}{3} 0 \frac{3}{3} 13 arccot (x) \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} arccot (x) 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ}

= \frac{π}{4} + πn

\frac{3 π + x}{2} \geq \frac{π}{4} + πn

Multiplicar ambos os lados por

2

\frac{3 π + x}{2} \geq \frac{π}{4} + πn

Multiplicar ambos os lados por

2

\frac{2 ( 3 π + x )}{2} \geq 2 \cdot \frac{π}{4} + 2 πn

Simplificar

\frac{2 ( 3 π + x )}{2} \geq 2 \cdot \frac{π}{4} + 2 πn

Simplificar

\frac{2 ( 3 π + x )}{2} : 3 π + x

\frac{2 ( 3 π + x )}{2}

Dividir:

\frac{2}{2} = 1

= 3 π + x

Simplificar

2 \cdot \frac{π}{4} + 2 πn : \frac{π}{2} + 2 πn

2 \cdot \frac{π}{4} + 2 πn

2 \cdot \frac{π}{4} = \frac{π}{2}

2 \cdot \frac{π}{4}

Multiplicar frações:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{π 2}{4}

Eliminar o fator comum:

2

= \frac{π}{2}

= \frac{π}{2} + 2 πn

3 π + x \geq \frac{π}{2} + 2 πn

3 π + x \geq \frac{π}{2} + 2 πn

3 π + x \geq \frac{π}{2} + 2 πn

Mova

3 π

para o lado direito

3 π + x \geq \frac{π}{2} + 2 πn

Subtrair

3 π

de ambos os lados

3 π + x - 3 π \geq \frac{π}{2} + 2 πn - 3 π

Simplificar

x \geq \frac{π}{2} + 2 πn - 3 π

x \geq \frac{π}{2} + 2 πn - 3 π

Simplificar

\frac{π}{2} - 3 π : - \frac{5 π}{2}

\frac{π}{2} - 3 π

Converter para fração:

3 π = \frac{3 π 2}{2}

= \frac{π}{2} - \frac{3 π 2}{2}

Já que os denominadores são iguais, combinar as frações:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{π - 3 π 2}{2}

π - 3 π 2 = - 5 π

π - 3 π 2

Multiplicar os números:

3 \cdot 2 = 6

= π - 6 π

Somar elementos similares:

π - 6 π = - 5 π

= - 5 π

= \frac{- 5 π}{2}

Aplicar as propriedades das frações:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{5 π}{2}

x \geq - \frac{5 π}{2} + 2 πn

\frac{3 π + x}{2} < π + πn : x < - π + 2 πn

\frac{3 π + x}{2} < π + πn

Multiplicar ambos os lados por

2

\frac{3 π + x}{2} < π + πn

Multiplicar ambos os lados por

2

\frac{2 ( 3 π + x )}{2} < 2 π + 2 πn

Simplificar

3 π + x < 2 π + 2 πn

3 π + x < 2 π + 2 πn

Mova

3 π

para o lado direito

3 π + x < 2 π + 2 πn

Subtrair

3 π

de ambos os lados

3 π + x - 3 π < 2 π + 2 πn - 3 π

Simplificar

x < - π + 2 πn

x < - π + 2 πn

Combinar os intervalos

x \geq - \frac{5 π}{2} + 2 πn and x < - π + 2 πn

Junte intervalos que se sobrepoem

- \frac{5 π}{2} + 2 πn \leq x < - π + 2 πn

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